340.5 g of sodium bromide To moles of sodium bromide

3.3 moles

Explanation:

Sodium bromide has a chemical formula of NaBr.

Given that:

Mass of NaBr in grams = 340.5g

Number of moles of NaBr = ?

For Molar mass of NaBr, use atomic mass of sodium Na = 23g, bromine Br = 79.9g

So,molar mass of NaBr = (23g + 79.9g)

= 102.9g/mol

Recall that number of moles = (mass in grams/molar mass)

No. of moles = 340.5g/102.9g/mol

= 3.3 moles

Thus, 340.5 g of sodium bromide gives 3.3 moles of sodium bromide

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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